通过bash脚本传递参数为/ bin / bash的 [英] Passing arguments to /bin/bash via a bash script

问题描述

我写一个bash脚本，通过登录shell取得了一些命令行参数（可能包括空格），并把他们都一个程序（如/ bin / some_program）。这是从bash脚本调用的登录shell将取决于用户的登录shell。让我们假设用户使用/斌/ bash作为在这个例子中登录shell ...但它可能是/斌/ tcsh的或其他任何东西。

如果我知道有多少参数将被传递给some_program，我可以把下面几行，我bash脚本：

 ＃！/斌/庆典
＃...（某​​些行，我们确定该用户的登录shell是bash）...
/斌/ bash的--login -c/斌/ some_program \\$ 1 \\\\$ 2 \\
 

，然后调用上面的脚本如下：

  my_script这太酷
 

通过上面的例子我可以证实，some_program接收两个参数，这太和酷。

我的问题是...如果我不知道有多少参数将被传递做什么呢？我想通过所有已发送到一起some_program my_script的参数。问题是我无法弄清楚如何做到这一点。这里有一些事情的不工作的：

  /斌/ bash的--login -c/斌/ some_program $ @＃ - ＆GT; 3个参数：这，是，过于
/斌/ bash的--login -c /斌/ some_program$ @＃ - ＆GT;通过无参数
 

解决方案

引用bash的手册 -c ：

  

如果有-c选项present，那么命令将从字符串读取。如果字符串后还有争论，他们被分配到位置参数，以$ 0起始

对我的作品：

  $猫x.sh
＃！/斌/庆典
/斌/ bash的--login -c'回声1：$ 1 2：$ 2 3：$ 3回响$ @
$ ./x.sh富巴，巴兹哎呀blargh quargh
1：富巴2：3巴兹：哎呀blargh quargh
 

我不知道你是怎么到达通过无参数的结论，也许你错过了 $ 1,0 位？

I am writing a bash script that takes a number of command line arguments (possibly including spaces) and passes all of them to a program (/bin/some_program) via a login shell. The login shell that is called from the bash script will depend on the user's login shell. Let's suppose the user uses /bin/bash as their login shell in this example... but it might be /bin/tcsh or anything else.

If I know how many arguments will be passed to some_program, I can put the following lines in my bash script:

#!/bin/bash
# ... (some lines where we determine that the user's login shell is bash) ...
/bin/bash --login -c "/bin/some_program \"$1\" \"$2\""

and then call the above script as follows:

my_script "this is too" cool

With the above example I can confirm that some_program receives two arguments, "this is too" and "cool".

My question is... what if I don't know how many arguments will be passed? I'd like to pass all the arguments that were sent to my_script along to some_program. The problem is I can't figure out how to do this. Here are some things that don't work:

/bin/bash --login -c "/bin/some_program $@"     # --> 3 arguments: "this","is","too"
/bin/bash --login -c /bin/some_program "$@"     # --> passes no arguments

解决方案

Quoting the bash manual for -c:

If the -c option is present, then commands are read from string. If there are arguments after the string, they are assigned to the positional parameters, starting with $0.

Works for me:

$ cat x.sh
#!/bin/bash
/bin/bash --login -c 'echo 1:$1 2:$2 3:$3' echo "$@"
$ ./x.sh "foo bar" "baz" "argh blargh quargh"
1:foo bar 2:baz 3:argh blargh quargh

I don't know how you arrived at the "passes no arguments" conclusion, maybe you missed the $0 bit?

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